Calculate Radius of Curvature for Involute of a Circle

[John O' Meara]

Show that for the involute of a circle (x=a(\cos\theta + \theta\sin\theta) \\, y=a(\sin\theta - \theta\cos\theta) for 0 <=\theta<=\pi \\) radius a, the radius of curvature, is \sqrt{2as}, where s is the arc length. The radius of curvature is \rho = \frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}\\ [\tex];<br /> My attempt: \frac{dx}{d\theta} = a \theta\cos\theta \\ and \frac{dy}{d\theta} = a\theta\sin\theta \\,<br /> that implies that \frac{dy}{dx} = \tan\theta \\, which implies that \frac{d^2y}{dx^2} = \sec^{2}\theta, therefore \rho = \sec\theta\\. Which is not the answer they suggest. Please help, thanks.<br /> <br />

 

[dextercioby]

Check again the second derivative of y wrt x. Be careful about using the chain rule. Remember that

\frac{d^2 y}{dx^2}=\frac{d}{d\Huge{\mathbf{x}}}\frac{dy}{dx}

 

[John O' Meara]

I checked \frac{d^2y}{dx^2} and got the following \frac{d^2y}{dx^2} = \frac{d\tan\theta}{d\theta} \cdot \frac{d\theta}{dx} = \sec^{2}\theta \cdot \frac{d\theta}{dx} = \frac{\sec^{2}\theta}{\frac{dx}{d\theta}} \\. This gives \frac{d^2y}{dx^2} = \frac{\sec^{2}\theta}{a\theta \cos\theta} \\, which yields \rho =a\theta. Which is not the correct answer either, but the integral of a\theta w.r.t, \theta gives s. Thanks for the reply.

 

[John O' Meara]

Is anyone interest as to why I can't get the correct answer. I get \rho = a\theta \mbox{ not the } \sqrt{2sa}? Thanks.

 

[dextercioby]

What is "s" equal to ? In terms of \rho,\theta, of course ?